Question

find the zeros of the polynomial and relationship between alpha and beta ​

Answer 1

Answer:

Step-by-step explanation:

Let p(x)= ax^{2}+bx+cp(x)=ax

2

+bx+c where a\neq 0a



​

=0 is a quadratic polynomial

\alpha, \betaα,β are the zeroes of p(x)p(x)

We know that

Sum of the roots = \alpha+ \beta = \dfrac{-b}{a}=α+β=

a

−b

​

Product of the roots = \alpha\times \beta = \dfrac{c}{a}=α× β=

a

c

​

We can also write p(x)p(x) in this form also

p(x)= x^{2}-\dfrac{-b}{a}x+\dfrac{c}{a}p(x)=x

2

−

a

−b

​

x+

a

c

​

\Rightarrow p(x)= x^{2}-(\alpha +\beta)x+\alpha \beta⇒p(x)=x

2

−(α+β)x+αβ

Now given that \alpha+ \beta=6α+β=6 and \alpha\beta =4αβ=4

Substituting these values we get

p(x)= x^{2}-(6)x+4p(x)=x

2

−(6)x+4 is the required polynomial.